Question

An observer is 28 m above the ground floor of a large hotel atrium looking at a​ glass-enclosed elevator shaft that is 26 m horizontally from the observer​ (see figure). The angle of elevation of the elevator is the angle that the​ observer's line of sight makes with the horizontal​ (it may be positive or​ negative). Assuming that the elevator rises at a rate of 6 m/s​, what is the rate of change of the angle of elevation when the elevator is 15 m above the​ ground? When the elevator is 54 m above the​ ground?

Answer 1

when the elevator is x meters up, then we have the angle of elevation θ is
tanθ = (x-28)/26 for 0<= x <= (some max height)
sec^2θ dθ/dt = 1/26 dx/dt
so, when x=15,
tanθ = -13/26 = -1/2
sec^2θ = 1 + tan^2θ = 5/4
and we have
5/4 dθ/dt = 1/26 * 6
dθ/dt = 12/65 rad/s

Answer 2

When the elevator is 54 m above the​ ground?

Answer 3

c'mon dude - use x=54
you gotta do <u>some</u> of the work!
Post your work if you get stuck.